How to solve calculations about logical address space and physical address space?

Question

How to calculate number of bits in logical address and physical address when logical address space of 8 pages of 1024 word each, mapped to physical memory of 32 frames?

Answer 1

15 is the correct answer

i think this is correct way size of logical address space is No. of pages * Page size = 8 * 1024 = 2^3 * 2 ^10 = 2^13 No. of bits for logical address is 13

Size of Physical address space is 2^5 * 2^10 = 2^15 No. of bits for physical address is 15

Answer 2

Consider the following room/floor analogy: Each floor in a hotel contains 10 rooms. The door in each room is labeled 01, 02, 03, ..., 10. Then you get off the elevator, there is a plaque with the floor number. There are 3 floors in this hotel: floors 1, 2, and 3. Therefore, you can say that, to eliminate the ambiguity in room numbers, you concatenate the floor number to the room in the following format: floor:room. So, 1:01 is different than 2:01, or 3:01.

Viewing this graphically:

1 | 01 | 02 | 03 | 04 | 05 | 06 | 07 | 08 | 09 | 10 |

2 | 01 | 02 | 03 | 04 | 05 | 06 | 07 | 08 | 09 | 10 |

3 | 01 | 02 | 03 | 04 | 05 | 06 | 07 | 08 | 09 | 10 |

The floor number can be expressed with one digit. The room number can be expressed with two digits. To express the unique location of the room (floor:room concatenation), you need three digits. Replace floor with frame, and room with page.